Hindawi Publishing Corporation
Advances in Mathematical Physics
Volume 2013, Article ID 423718, 7 pages
http://dx.doi.org/10.1155/2013/423718
Research Article
Conservative Linear Difference Scheme for
Rosenau-KdV Equation
Jinsong Hu,1 Youcai Xu,2 and Bing Hu2
1
2
School of Mathematics and Computer Engineering, Xihua University, Chengdu 610039, China
School of Mathematics, Sichuan University, Chengdu 610064, China
Correspondence should be addressed to Youcai Xu; xyc@scu.edu.cn
Received 5 February 2013; Accepted 22 March 2013
Academic Editor: Hagen Neidhardt
Copyright © 2013 Jinsong Hu et al. This is an open access article distributed under the Creative Commons Attribution License,
which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
A conservative three-level linear finite difference scheme for the numerical solution of the initial-boundary value problem of
Rosenau-KdV equation is proposed. The difference scheme simulates two conservative quantities of the problem well. The existence
and uniqueness of the difference solution are proved. It is shown that the finite difference scheme is of second-order convergence
and unconditionally stable. Numerical experiments verify the theoretical results.
1. Introduction
KdV equation has been used in very wide applications and
undergone research which can be used to describe wave
propagation and spread interaction as follows [1–4]:
𝑢𝑡 + 𝑢𝑢𝑥 + 𝑢𝑥𝑥𝑥 = 0.
(1)
In the study of the dynamics of dense discrete systems, the
case of wave-wave and wave-wall interactions cannot be
described using the well-known KdV equation. To overcome
this shortcoming of the KdV equation, Rosenau [5, 6]
proposed the so-called Rosenau equation:
𝑢𝑡 + 𝑢𝑥𝑥𝑥𝑥𝑡 + 𝑢𝑥 + 𝑢𝑢𝑥 = 0.
(2)
The existence and the uniqueness of the solution for (2) were
proved by Park [7], but it is difficult to find the analytical
solution for (2). Since then, much work has been done on
the numerical method for (2) ([8–13] and also the references
therein). On the other hand, for the further consideration
of the nonlinear wave, the viscous term +𝑢𝑥𝑥𝑥 needs to be
included [14]
𝑢𝑡 + 𝑢𝑥𝑥𝑥𝑥𝑡 + 𝑢𝑥 + 𝑢𝑢𝑥 + 𝑢𝑥𝑥𝑥 = 0.
(3)
This equation is usually called the Rosenau-KdV equation.
Zuo [14] discussed the solitary wave solutions and periodic
solutions for (2). Recently, [15–17] discussed the solitary solutions for the generalized Rosenau-KdV equation with usual
power law nonlinearity. In [15, 16], the authors also gave the
two invariants for the generalized Rosenau-KdV equation.
In particular, [16] not only derived the singular 1-solition
solution by the ansatz method but also used perturbation
theory to obtain the adiabatic parameter dynamics of the
water waves. In [17], The ansatz method is applied to obtain
the topological soliton solution of the generalized RosenauKdV equation. The 𝐺󸀠 /𝐺 method as well as the exp-function
method are also applied to extract a few more solutions
to this equation. But the numerical method to the initialboundary value problem of Rosenau-KdV equation has not
been studied till now. In this paper, we propose a conservative
three-level finite difference scheme for the Rosenau-KdV
equation (3) with the boundary conditions
𝑢 (𝑥𝐿 , 𝑡) = 𝑢 (𝑥𝑅 , 𝑡) = 0,
𝑢𝑥 (𝑥𝐿 , 𝑡) = 𝑢𝑥 (𝑥𝑅 , 𝑡) = 0,
𝑢𝑥𝑥 (𝑥𝐿 , 𝑡) = 𝑢𝑥𝑥 (𝑥𝑅 , 𝑡) = 0,
𝑡 ∈ [0, 𝑇] ,
(4)
and an initial condition
𝑢 (𝑥, 0) = 𝑢0 (𝑥) ,
𝑥 ∈ [𝑥𝐿 , 𝑥𝑅 ] .
(5)
2
Advances in Mathematical Physics
The initial boundary value problem (3)–(5) possesses the
following conservative properties [15]:
𝑥𝑅
of ℎ and 𝜏, which may have different values in different
occurrences. We introduce the following notations:
(𝑢𝑗𝑛 )
𝑥
𝑥𝑅
𝑄 (𝑡) = ∫ 𝑢 (𝑥, 𝑡) 𝑑𝑥 = ∫ 𝑢0 (𝑥) 𝑑𝑥 = 𝑄 (0) ,
𝑥𝐿
𝑥𝐿
󵄩 󵄩2
𝐸 (𝑡) = ‖𝑢‖2𝐿 2 + 󵄩󵄩󵄩𝑢𝑥𝑥 󵄩󵄩󵄩𝐿 2 = 𝐸 (0) .
(6)
(𝑢𝑗𝑛 ) ̂ =
𝑢𝑛𝑗
1√
−26 + 2√313
24
𝑛
𝑛
𝑢𝑗+1
− 𝑢𝑗−1
2ℎ
=
𝑢𝑗𝑛+1 + 𝑢𝑗𝑛−1
2
(𝑢𝑗𝑛 )
𝑥
,
=
𝑛
𝑢𝑗𝑛 − 𝑢𝑗−1
𝑛
,
𝑢𝑗𝑛+1 − 𝑢𝑗𝑛−1
2𝜏
𝑡
⟨𝑢 , V ⟩ =
,
ℎ
(𝑢𝑗𝑛 )̂ =
,
𝑛
,
𝐽−1
(9)
ℎ ∑ 𝑢𝑗𝑛 V𝑗𝑛 ,
𝑗=1
󵄩 𝑛󵄩
󵄩󵄩 𝑛 󵄩󵄩
󵄩󵄩𝑢 󵄩󵄩∞ = max 󵄩󵄩󵄩󵄩𝑢𝑗 󵄩󵄩󵄩󵄩 .
1≤𝑗≤𝐽−1
󵄩󵄩 𝑛 󵄩󵄩2
𝑛 𝑛
󵄩󵄩𝑢 󵄩󵄩 = ⟨𝑢 , 𝑢 ⟩ ,
35
35 √
313)
+
24 312
× sech4 [
ℎ
𝑥
(7)
The solitary wave solution for (3) is [14, 15]
𝑢 (𝑥, 𝑡) = (−
=
𝑛
𝑢𝑗+1
− 𝑢𝑗𝑛
We propose a three-level linear finite difference scheme
for the solution of (3)–(5) as follows:
(8)
(𝑢𝑗𝑛 )̂ + (𝑢𝑗𝑛 )
𝑡
1
1
× (𝑥 − ( + √313) 𝑡) ] .
2 26
𝑥𝑥𝑥𝑥̂𝑡
+ (𝑢𝑛𝑗 ) ̂ + (𝑢𝑛𝑗 )
𝑥
1
+ [𝑢𝑗𝑛 (𝑢𝑛𝑗 ) ̂ + (𝑢𝑗𝑛 𝑢𝑛𝑗 ) ̂ ] = 0,
𝑥
𝑥
3
𝑗 = 1, 2, 3, . . . , 𝐽 − 1,
When −𝑥𝐿 ≫ 0, 𝑥𝑅 ≫ 0, the initial-boundary value problem
(3)–(5) and the Cauchy problem (3) are consistent, so the
boundary condition (4) is reasonable.
It is known the conservative scheme is better than the
nonconservative ones. The nonconservative scheme may
easily show nonlinear blow up. A lot of numerical experiments show that the conservative scheme can possesses
some invariant properties of the original differential equation
[18–29]. The conservative scheme is more suitable for longtime calculations. In [19], Li and Vu-Quoc said “. . . in some
areas, the ability to preserve some invariant properties of
the original differential equation is a criterion to judge the
success of a numerical simulation.” In this paper, we propose
a three-level linear finite difference scheme for the RosenauKdV equation (3)–(5). The difference scheme is conservative
which simulates conservative properties (6) and (7) at the
same time.
The rest of this paper is organized as follows. In Section 2,
we propose a three-level linear finite difference scheme for the
Rosenau-KdV equation and discuss the discrete conservative
properties. In Section 3, we show that the scheme is uniquely
solvable. Then, in Section 4, we prove that the finite difference
scheme is of second-order convergence, unconditionally
stable. Finally, some numerical tests are given in Section 5 to
verify our theoretical analysis.
2. Finite Difference Scheme and
Conservation Properties
Let ℎ = (𝑥𝑅 −𝑥𝐿 )/𝐽 and 𝜏 be the uniform step size in the spatial
and temporal direction, respectively. Denote 𝑥𝑗 = 𝑥𝐿 +𝑗ℎ (𝑗 =
−1, 0, 1, 2, . . . , 𝐽, 𝐽 + 1), 𝑡𝑛 = 𝑛𝜏 (𝑛 = 0, 1, 2, . . . , 𝑁, 𝑁 =
[𝑇/𝜏]), 𝑢𝑗𝑛 ≈ 𝑢(𝑥𝑗 , 𝑡𝑛 ) and 𝑍ℎ0 = {𝑢 = (𝑢𝑗 ) | 𝑢−1 = 𝑢0 =
𝑢𝐽 = 𝑢𝐽+1 = 0, 𝑗 = −1, 0, 1, 2, . . . , 𝐽, 𝐽 + 1}. Throughout this
paper, we denote 𝐶 as a generic positive constant independent
𝑥𝑥𝑥̂
𝑢𝑗0 = 𝑢0 (𝑥𝑗 ) ,
𝑢𝑛 ∈ 𝑍ℎ0 ,
𝑛 = 1, 2, 3, . . . , 𝑁 − 1,
𝑗 = 0, 1, 2, 3, . . . , 𝐽,
(𝑢0𝑛 )𝑥̂ = (𝑢𝐽𝑛 )𝑥̂ = 0,
(𝑢0𝑛 )𝑥𝑥 = (𝑢𝐽𝑛 )𝑥𝑥 = 0,
𝑛 = 1, 2, 3, . . . , 𝑁.
(10)
(11)
(12)
(13)
From the boundary conditions (4), we know that (13) is
reasonable.
Lemma 1. It follows from summation by parts that for any two
mesh functions 𝑢, V ∈ 𝑍ℎ0 ,
⟨𝑢𝑥 , V⟩ = − ⟨𝑢, V𝑥 ⟩ ,
⟨𝑢𝑥𝑥 , V⟩ = − ⟨𝑢𝑥 , V𝑥 ⟩ .
(14)
Then one has
󵄩 󵄩2
⟨𝑢𝑥𝑥 , 𝑢⟩ = − ⟨𝑢𝑥 , 𝑢𝑥 ⟩ = −󵄩󵄩󵄩𝑢𝑥 󵄩󵄩󵄩 .
(15)
⟨𝑢𝑥 , 𝑢⟩ = − ⟨𝑢, 𝑢𝑥 ⟩ ,
Furthermore, if (𝑢0 )𝑥𝑥 = (𝑢𝐽 )𝑥𝑥 = 0, then
󵄩 󵄩2
⟨𝑢𝑥𝑥𝑥𝑥 , 𝑢⟩ = 󵄩󵄩󵄩𝑢𝑥𝑥 󵄩󵄩󵄩 .
(16)
The difference scheme (10)–(13) simulates two conservative properties of the problems (6) and (7) as follows.
Theorem 2. Suppose that 𝑢0 ∈ 𝐻02 [𝑥𝐿 , 𝑥𝑅 ], 𝑢(𝑥, 𝑡) ∈ 𝐶5,3 ,
then the difference scheme (10)–(13) is conservative:
𝑄𝑛 =
ℎ 𝐽−1 𝑛+1
ℎ 𝐽−1
∑ (𝑢𝑗 + 𝑢𝑗𝑛 ) + 𝜏 ∑ 𝑢𝑗𝑛 (𝑢𝑗𝑛+1 ) ̂ = 𝑄𝑛−1 = ⋅ ⋅ ⋅ = 𝑄0 ,
𝑥
2 𝑗=1
6 𝑗=1
(17)
𝐸𝑛 =
1 󵄩󵄩 𝑛+1 󵄩󵄩2 󵄩󵄩 𝑛 󵄩󵄩2
1 󵄩 𝑛+1 󵄩󵄩2 󵄩󵄩 𝑛 󵄩󵄩2
󵄩󵄩 + 󵄩󵄩𝑢𝑥𝑥 󵄩󵄩 )
(󵄩󵄩󵄩𝑢 󵄩󵄩󵄩 + 󵄩󵄩𝑢 󵄩󵄩 ) + (󵄩󵄩󵄩󵄩𝑢𝑥𝑥
󵄩
2
2
=𝐸
𝑛−1
0
= ⋅⋅⋅ = 𝐸 .
(18)
Advances in Mathematical Physics
3
Proof. Multiplying (10) with ℎ, summing up for 𝑗 from 1 to 𝐽−
1, and considering the boundary condition (13) and Lemma 1,
we get
𝐽−1 𝑢𝑛+1
𝑗
− 𝑢𝑗𝑛−1
𝑗=1
2𝜏
ℎ∑
+
ℎ 𝐽−1 𝑛 𝑛+1
∑ [𝑢 (𝑢 ) − 𝑢𝑗𝑛−1 (𝑢𝑗𝑛 ) ̂ ] = 0.
𝑥
6 𝑗=1 𝑗 𝑗 𝑥̂
We also can get 𝑢1 in order 𝑂(ℎ2 + 𝜏2 ) by two-level 𝐶-𝑁
scheme (i.e., 𝑢0 and 𝑢1 are uniquely determined). Now
suppose 𝑢0 , 𝑢1 , . . . , 𝑢𝑛 (1 ≤ 𝑛 ≤ 𝑁 − 1) is solved uniquely.
Consider the equation of (10) for 𝑢𝑛+1 :
1 𝑛+1 1 𝑛+1
1
+ (𝑢𝑗𝑛+1 ) ̂
𝑢𝑗 + (𝑢𝑗 )
𝑥𝑥𝑥𝑥
𝑥
2𝜏
2𝜏
2
(19)
Then, (17) is gotten from (19).
Taking an inner product of (10) with 2𝑢𝑛 (i.e., 𝑢𝑛+1 +𝑢𝑛−1 ),
considering the boundary condition (13) and Lemma 1, we
obtain
Taking an inner product of (24) with 𝑢𝑛+1 , we get
1 󵄩󵄩 𝑛+1 󵄩󵄩2 1 󵄩󵄩 𝑛+1 󵄩󵄩2 1 𝑛+1 𝑛+1
1 𝑛+1 𝑛+1
󵄩󵄩𝑢 󵄩󵄩 + 󵄩󵄩𝑢𝑥𝑥 󵄩󵄩 + ⟨𝑢𝑥̂ , 𝑢 ⟩ + ⟨𝑢𝑥𝑥
⟩
𝑥̂ , 𝑢
󵄩
󵄩
󵄩
󵄩
2𝜏
2𝜏
2
2
1 󵄩󵄩 𝑛+1 󵄩󵄩2 󵄩󵄩 𝑛−1 󵄩󵄩2
1 󵄩󵄩 𝑛+1 󵄩󵄩2 󵄩󵄩 𝑛−1 󵄩󵄩2
(󵄩󵄩𝑢 󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑢 󵄩󵄩󵄩 ) +
(󵄩󵄩𝑢 󵄩󵄩 − 󵄩󵄩𝑢 󵄩󵄩 )
2𝜏 󵄩
2𝜏 󵄩 𝑥𝑥 󵄩 󵄩 𝑥𝑥 󵄩
+ 2 ⟨𝑢𝑛𝑥̂ , 𝑢𝑛 ⟩ + 2 ⟨𝑢𝑛𝑥𝑥𝑥̂ , 𝑢𝑛 ⟩
1
1
+ (𝑢𝑗𝑛+1 ) ̂ + [𝑢𝑗𝑛 (𝑢𝑗𝑛+1 ) ̂ + (𝑢𝑗𝑛 𝑢𝑗𝑛+1 ) ̂ ] = 0.
𝑥𝑥𝑥
𝑥
𝑥
2
6
(24)
(20)
+
+ 2 ⟨𝑃, 𝑢𝑛 ⟩ = 0,
(25)
where 𝑃𝑗 = (1/3)[𝑢𝑗𝑛 (𝑢𝑛𝑗 )𝑥̂ + (𝑢𝑗𝑛 𝑢𝑛𝑗 )𝑥̂ ]. According to
⟨𝑢𝑛𝑥̂ , 𝑢𝑛 ⟩
Similar to the proof of (21), we have
= 0,
𝑛+1 𝑛+1
⟩ = 0.
⟨𝑢𝑥𝑥
𝑥̂ , 𝑢
1 𝐽−1
⟨𝑃, 𝑢𝑛 ⟩ = ℎ ∑ [𝑢𝑗𝑛 (𝑢𝑛𝑗 ) ̂ + (𝑢𝑗𝑛 𝑢𝑛𝑗 ) ̂ ] 𝑢𝑛𝑗
𝑥
𝑥
3 𝑗=1
ℎ 𝐽−1 𝑛 𝑛+1
∑ [𝑢 (𝑢 ) + (𝑢𝑗𝑛 𝑢𝑗𝑛+1 ) ̂ ] 𝑢𝑗𝑛+1
𝑥
6 𝑗=1 𝑗 𝑗 𝑥̂
1
𝑛−1
𝑛+1
𝑛−1
− 𝑢𝑗−1
− 𝑢𝑗−1
)
∑ [𝑢𝑛 (𝑢𝑛+1 + 𝑢𝑗+1
12 𝑗=1 𝑗 𝑗+1
𝑛+1
(𝑢𝑗+1
+
𝑛−1
𝑢𝑗+1
)
−
𝑛
𝑢𝑗−1
𝑛+1
(𝑢𝑗−1
+
𝑛−1
𝑢𝑗−1
)]
× (𝑢𝑗𝑛+1 + 𝑢𝑗𝑛−1 )
=
1 𝐽−1 𝑛
𝑛
𝑛+1
𝑛−1
) (𝑢𝑗+1
+ 𝑢𝑗+1
) (𝑢𝑗𝑛+1 + 𝑢𝑗𝑛−1 )
∑ (𝑢 + 𝑢𝑗+1
12 𝑗=1 𝑗
−
(26)
By
𝐽−1
𝑛
+𝑢𝑗+1
𝑛+1
⟩ = 0,
⟨𝑢𝑥𝑛+1
̂ ,𝑢
(21)
⟨𝑢𝑛𝑥𝑥𝑥̂ , 𝑢𝑛 ⟩ = 0,
=
ℎ 𝐽−1 𝑛 𝑛+1
∑ [𝑢 (𝑢 ) + (𝑢𝑗𝑛 𝑢𝑗𝑛+1 ) ̂ ] 𝑢𝑗𝑛+1 = 0.
𝑥
6 𝑗=1 𝑗 𝑗 𝑥̂
1 𝐽−1 𝑛
𝑛
𝑛+1
𝑛−1
) (𝑢𝑗𝑛+1 + 𝑢𝑗𝑛−1 ) (𝑢𝑗−1
+ 𝑢𝑗−1
)
∑ (𝑢 + 𝑢𝑗−1
12 𝑗=1 𝑗
1 𝐽−1 𝑛 𝑛+1
𝑛+1
𝑛
𝑛+1
𝑛
𝑛+1
) + (𝑢𝑗+1
𝑢𝑗+1
− 𝑢𝑗−1
𝑢𝑗−1
)] 𝑢𝑗𝑛+1
∑ [𝑢 (𝑢 − 𝑢𝑗−1
12 𝑗=1 𝑗 𝑗+1
=
1 𝐽−1 𝑛 𝑛+1 𝑛+1
𝑛
𝑛+1
𝑢𝑗𝑛+1 𝑢𝑗+1
]
∑ [𝑢 𝑢 𝑢 + 𝑢𝑗+1
12 𝑗=1 𝑗 𝑗 𝑗+1
−
1 𝐽−1 𝑛 𝑛+1 𝑛+1
𝑛+1 𝑛+1
𝑢𝑗 ] = 0,
∑ [𝑢 𝑢 𝑢 + 𝑢𝑗𝑛 𝑢𝑗−1
12 𝑗=1 𝑗−1 𝑗−1 𝑗
(27)
and from (25)–(27), we have
= 0,
(22)
we have
󵄩
󵄩2 󵄩
󵄩2
󵄩 𝑛+1 󵄩󵄩2 󵄩󵄩 𝑛−1 󵄩󵄩2
󵄩󵄩 − 󵄩󵄩𝑢𝑥𝑥 󵄩󵄩 ) = 0.
(󵄩󵄩󵄩󵄩𝑢𝑛+1 󵄩󵄩󵄩󵄩 − 󵄩󵄩󵄩󵄩𝑢𝑛−1 󵄩󵄩󵄩󵄩 ) + (󵄩󵄩󵄩󵄩𝑢𝑥𝑥
󵄩 󵄩
󵄩
=
(23)
󵄩󵄩 𝑛+1 󵄩󵄩2 󵄩󵄩 𝑛+1 󵄩󵄩2
󵄩󵄩𝑢 󵄩󵄩 + 󵄩󵄩𝑢𝑥𝑥 󵄩󵄩 = 0.
󵄩
󵄩 󵄩
󵄩
(28)
That is, (24) has only a trivial solution. Therefore, (10)
determines 𝑢𝑗𝑛+1 uniquely. This completes the proof.
Then, (18) is gotten from (23).
4. Convergence and Stability
3. Solvability
Let V(𝑥, 𝑡) be the solution of problem (3)–(5), V𝑗𝑛 = V(𝑥𝑗 , 𝑡𝑛 ),
then the truncation error of the difference scheme (10)–(13)
is as follows:
Theorem 3. There exists 𝑢𝑛 ∈ 𝑍ℎ0 which satisfies the difference
scheme (10)–(13), (1 ≤ 𝑛 ≤ 𝑁).
Proof. Use mathematical induction to prove it. It is obvious
that 𝑢0 is uniquely determined by the initial condition (12).
𝑟𝑗𝑛 = (V𝑗𝑛 )̂ + (V𝑗𝑛 )
𝑥𝑥𝑥𝑥̂𝑡
𝑡
+ (V𝑛𝑗 )
+
̂
𝑥𝑥𝑥
+ (V𝑛𝑗 ) ̂
𝑥
1 𝑛 𝑛
[V (V ) + (V𝑗𝑛 V𝑛𝑗 ) ̂ ] .
𝑥
3 𝑗 𝑗 𝑥̂
(29)
4
Advances in Mathematical Physics
Making use of Taylor expansion, we know that 𝑟𝑗𝑛 = 𝑂(𝜏2 +ℎ2 )
holds if ℎ, 𝜏 → 0.
Lemma 4. Suppose that 𝑢0 ∈ 𝐻02 [𝑥𝐿 , 𝑥𝑅 ], then the solution 𝑢𝑛
of (3)–(5) satisfies
‖𝑢‖𝐿 2 ≤ 𝐶,
‖𝑢‖𝐿 ∞ ≤ 𝐶,
󵄩󵄩 󵄩󵄩
󵄩󵄩𝑢𝑥 󵄩󵄩𝐿 2 ≤ 𝐶,
󵄩󵄩󵄩𝑢𝑥 󵄩󵄩󵄩 ≤ 𝐶.
󵄩 󵄩𝐿 ∞
(30)
Theorem 8. Suppose 𝑢0 ∈ 𝐻02 [𝑥𝐿 , 𝑥𝑅 ], 𝑢(𝑥, 𝑡) ∈ 𝐶5,3 , then the
solution 𝑢𝑛 of the difference scheme (10)–(13) converges to the
solution V(𝑥, 𝑡) of the problem (3)–(5) with order 𝑂(𝜏2 + ℎ2 ) in
norm ‖ ⋅ ‖∞ .
Proof. Subtracting (10) from (29) and letting 𝑒𝑗𝑛 = V𝑗𝑛 − 𝑢𝑗𝑛 , we
have
𝑟𝑗𝑛 = (𝑒𝑗𝑛 )̂ + (𝑒𝑛𝑗 )
𝑡
𝑥𝑥𝑥𝑥̂𝑡
+ (𝑒𝑛𝑗 ) ̂ + (𝑒𝑛𝑗 )
𝑥
𝑥𝑥𝑥̂
+ 𝑅1,𝑗 + 𝑅2,𝑗 ,
Proof. It is follows from (7) that
‖𝑢‖𝐿 2 ≤ 𝐶,
󵄩󵄩 󵄩󵄩
󵄩󵄩𝑢𝑥𝑥 󵄩󵄩𝐿 2 ≤ 𝐶.
(31)
By Holder inequality and Schwarz inequality, we get
𝑥
𝑥
𝑥𝐿
𝑥𝐿
𝑅
𝑅
󵄨𝑥
󵄩󵄩 󵄩󵄩2
󵄩󵄩𝑢𝑥 󵄩󵄩𝐿 2 = ∫ 𝑢𝑥 𝑢𝑥 𝑑𝑥 = 𝑢𝑢𝑥 󵄨󵄨󵄨𝑥𝑅𝐿 − ∫ 𝑢𝑢𝑥𝑥 𝑑𝑥
𝑥𝑅
= − ∫ 𝑢𝑢𝑥𝑥 𝑑𝑥
where 𝑅1,𝑗 = (1/3)[V𝑗𝑛 (V𝑛𝑗 )𝑥̂ − 𝑢𝑗𝑛 (𝑢𝑛𝑗 )𝑥̂ ], 𝑅2,𝑗 = (1/3)[(V𝑗𝑛 V𝑛𝑗 )𝑥̂ −
(𝑢𝑗𝑛 𝑢𝑛𝑗 )𝑥̂ ]. Computing the inner product of (39) with 2𝑒𝑛 , we
obtain
⟨𝑟𝑛 , 2𝑒𝑛 ⟩ =
(32)
𝑥𝐿
1
󵄩 󵄩
󵄩 󵄩2
≤ ‖𝑢‖𝐿 2 ⋅ 󵄩󵄩󵄩𝑢𝑥𝑥 󵄩󵄩󵄩𝐿 2 ≤ (‖𝑢‖2𝐿 2 + 󵄩󵄩󵄩𝑢𝑥𝑥 󵄩󵄩󵄩𝐿 2 ) ,
2
(39)
1 󵄩󵄩 𝑛+1 󵄩󵄩2 󵄩󵄩 𝑛−1 󵄩󵄩2
1 󵄩󵄩 𝑛+1 󵄩󵄩2 󵄩󵄩 𝑛−1 󵄩󵄩2
(󵄩󵄩𝑒 󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑒 󵄩󵄩󵄩 ) +
(󵄩󵄩𝑒 󵄩󵄩 − 󵄩󵄩𝑒 󵄩󵄩 )
2𝜏 󵄩
2𝜏 󵄩 𝑥𝑥 󵄩 󵄩 𝑥𝑥 󵄩
+ ⟨𝑒𝑛𝑥̂ , 2𝑒𝑛 ⟩ + ⟨𝑒𝑛𝑥𝑥𝑥̂ , 2𝑒𝑛 ⟩ + ⟨𝑅1 , 2𝑒𝑛 ⟩ + ⟨𝑅2 , 2𝑒𝑛 ⟩ .
(40)
Similar to the proof of (21), we have
which implies that
󵄩󵄩 󵄩󵄩
󵄩󵄩𝑢𝑥 󵄩󵄩𝐿 2 ≤ 𝐶.
⟨𝑒𝑛𝑥̂ , 2𝑒𝑛 ⟩ = 0,
(33)
Using Sobolev inequality, we get that ‖𝑢‖𝐿 ∞ ≤ 𝐶, ‖𝑢𝑥 ‖𝐿 ∞ ≤
𝐶.
Lemma 5 (discrete Sobolev’s inequality [27]). There exist two
constants 𝐶1 and 𝐶2 such that
󵄩󵄩 𝑛 󵄩󵄩
󵄩 𝑛󵄩
󵄩 𝑛󵄩
󵄩󵄩𝑢 󵄩󵄩∞ ≤ 𝐶1 󵄩󵄩󵄩𝑢 󵄩󵄩󵄩 + 𝐶2 󵄩󵄩󵄩𝑢𝑥 󵄩󵄩󵄩 .
(34)
Lemma 6 (discrete Gronwall inequality [27]). Suppose that
𝑤(𝑘) and 𝜌(𝑘) are nonnegative function and 𝜌(𝑘) is nondecreasing. If 𝐶 > 0, and
𝑘−1
𝑤 (𝑘) ≤ 𝜌 (𝑘) + 𝐶𝜏 ∑ 𝑤 (𝑙) ,
∀𝑘,
(35)
𝑙=0
⟨𝑒𝑛𝑥𝑥𝑥̂ , 2𝑒𝑛 ⟩ = 0.
(41)
Then, (40) can be rewritten as follows:
󵄩2 󵄩
󵄩2
󵄩 𝑛+1 󵄩󵄩2 󵄩󵄩 𝑛−1 󵄩󵄩2
󵄩
󵄩󵄩 − 󵄩󵄩𝑒𝑥𝑥 󵄩󵄩 )
(󵄩󵄩󵄩󵄩𝑒𝑛+1 󵄩󵄩󵄩󵄩 − 󵄩󵄩󵄩󵄩𝑒𝑛+1 󵄩󵄩󵄩󵄩 ) + (󵄩󵄩󵄩󵄩𝑒𝑥𝑥
󵄩 󵄩
󵄩
= 2𝜏 ⟨𝑟𝑛 , 2𝑒𝑛 ⟩ + 2𝜏 ⟨−𝑅1 , 2𝑒𝑛 ⟩ + 2𝜏 ⟨−𝑅2 , 2𝑒𝑛 ⟩ .
(42)
Using Lemma 4 and Theorem 7, we get
󵄨󵄨 𝑛 󵄨󵄨
󵄨󵄨(𝑢 ) 󵄨󵄨 ≤ 𝐶,
󵄨󵄨 𝑗 𝑥̂ 󵄨󵄨
󵄨󵄨 𝑛 󵄨󵄨
󵄨󵄨V𝑗 󵄨󵄨 ≤ 𝐶,
󵄨 󵄨
󵄨󵄨 𝑛 󵄨󵄨
󵄨󵄨𝑢𝑗 󵄨󵄨 ≤ 𝐶,
󵄨 󵄨
(𝑗 = 0, 1, 2, . . . , 𝐽; 𝑛 = 1, 2, . . . , 𝑁) .
(43)
then
𝑤 (𝑘) ≤ 𝜌 (𝑘) 𝑒𝐶𝜏𝑘 ,
∀𝑘.
(36)
Theorem 7. Suppose 𝑢0 ∈ 𝐻02 [𝑥𝐿 , 𝑥𝑅 ], then the solution of
𝑛
‖ ≤ 𝐶, which
(10)–(13) satisfies: ‖𝑢𝑛 ‖ ≤ 𝐶, ‖𝑢𝑥𝑛 ‖ ≤ 𝐶, ‖𝑢𝑥𝑥
𝑛
𝑛
yield ‖𝑢 ‖∞ ≤ 𝐶, ‖𝑢𝑥 ‖∞ ≤ 𝐶 (𝑛 = 1, 2, . . . , 𝑁).
Proof. It is follows from (18) that
󵄩󵄩 𝑛 󵄩󵄩
󵄩󵄩𝑢 󵄩󵄩 ≤ 𝐶,
󵄩󵄩 𝑛 󵄩󵄩
󵄩󵄩𝑢𝑥𝑥 󵄩󵄩 ≤ 𝐶.
(37)
According to (15) and Schwarz inequality, we get
󵄩󵄩 𝑛 󵄩󵄩2 󵄩󵄩 𝑛 󵄩󵄩 󵄩󵄩 𝑛 󵄩󵄩 1 󵄩󵄩 𝑛 󵄩󵄩2 󵄩󵄩 𝑛 󵄩󵄩2
󵄩󵄩𝑢𝑥 󵄩󵄩 ≤ 󵄩󵄩𝑢 󵄩󵄩 ⋅ 󵄩󵄩𝑢𝑥𝑥 󵄩󵄩 ≤ (󵄩󵄩𝑢 󵄩󵄩 + 󵄩󵄩𝑢𝑥𝑥 󵄩󵄩 ) ≤ 𝐶.
2
Using Lemma 5, we have ‖𝑢𝑛 ‖∞ ≤ 𝐶, ‖𝑢𝑥𝑛 ‖∞ ≤ 𝐶.
(38)
According to the Schwarz inequality, we obtain
⟨−𝑅1 , 2𝑒𝑛 ⟩
2 𝐽−1
= − ℎ ∑ [V𝑗𝑛 (V𝑛𝑗 ) ̂ − 𝑢𝑗𝑛 (𝑢𝑛𝑗 ) ̂ ] 𝑒𝑛𝑗
𝑥
𝑥
3 𝑗=1
2 𝐽−1
= − ℎ ∑ [V𝑗𝑛 (𝑒𝑛𝑗 ) ̂ + 𝑒𝑗𝑛 (𝑢𝑛𝑗 ) ̂ ] 𝑒𝑛𝑗
𝑥
𝑥
3 𝑗=1
2 𝐽−1 󵄨󵄨
󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2
󵄨󵄨 󵄨 󵄨 󵄨 󵄨
≤ 𝐶ℎ ∑ (󵄨󵄨󵄨(𝑒𝑛𝑗 ) ̂ 󵄨󵄨󵄨 + 󵄨󵄨󵄨󵄨𝑒𝑗𝑛 󵄨󵄨󵄨󵄨) 󵄨󵄨󵄨󵄨𝑒𝑛𝑗 󵄨󵄨󵄨󵄨 ≤ 𝐶 [󵄩󵄩󵄩𝑒𝑛𝑥 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑒𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑒𝑛 󵄩󵄩󵄩 ]
𝑥
󵄨
󵄨
3 𝑗=1
󵄩2 󵄩
󵄩2 󵄩 󵄩2 󵄩
󵄩2 󵄩
󵄩2
󵄩
≤ 𝐶 [󵄩󵄩󵄩󵄩𝑒𝑛+1 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩𝑒𝑛−1 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑒𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩𝑒𝑥𝑛+1 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩𝑒𝑥𝑛−1 󵄩󵄩󵄩󵄩 ] ,
Advances in Mathematical Physics
5
⟨−𝑅2 , 2𝑒𝑛 ⟩
which yields
󵄩 󵄩2
(1 − 𝐶𝜏) (𝐵𝑛 − 𝐵𝑛−1 ) ≤ 2𝐶𝜏𝐵𝑛−1 + 2𝜏󵄩󵄩󵄩𝑟𝑛 󵄩󵄩󵄩 .
2 𝐽−1
= − ℎ ∑ [(V𝑗𝑛 V𝑛𝑗 ) ̂ − (𝑢𝑗𝑛 𝑢𝑛𝑗 ) ̂ ] 𝑒𝑛𝑗
𝑥
𝑥
3 𝑗=1
(50)
If 𝜏 is sufficiently small, which satisfies 1 − 𝐶𝜏 > 0, then
𝐽−1
2
= ℎ ∑ [V𝑗𝑛 V𝑛𝑗 − 𝑢𝑗𝑛 𝑢𝑛𝑗 ] (𝑒𝑛𝑗 ) ̂
𝑥
3 𝑗=1
󵄩 󵄩2
𝐵𝑛 − 𝐵𝑛−1 ≤ 𝐶𝜏𝐵𝑛−1 + 𝐶𝜏󵄩󵄩󵄩𝑟𝑛 󵄩󵄩󵄩 .
(51)
Summing up (51) from 1 to 𝑛, we have
𝐽−1
2
= ℎ ∑ [V𝑗𝑛 𝑒𝑛𝑗 + 𝑒𝑗𝑛 𝑢𝑛𝑗 ] (𝑒𝑛𝑗 ) ̂
𝑥
3 𝑗=1
𝑛
𝑛−1
󵄩 󵄩2
𝐵𝑛 ≤ 𝐵0 + 𝐶𝜏∑󵄩󵄩󵄩󵄩𝑟𝑙 󵄩󵄩󵄩󵄩 + 𝐶𝜏 ∑ 𝐵𝑙 .
𝑙=1
󵄨󵄨
2 𝐽−1 󵄨
󵄨 󵄨 󵄨 󵄨󵄨
≤ 𝐶ℎ ∑ (󵄨󵄨󵄨󵄨(𝑒𝑛𝑗 )󵄨󵄨󵄨󵄨 + 󵄨󵄨󵄨󵄨𝑒𝑗𝑛 󵄨󵄨󵄨󵄨) 󵄨󵄨󵄨(𝑒𝑛𝑗 ) ̂ 󵄨󵄨󵄨
𝑥
󵄨
󵄨
3 𝑗=1
(52)
𝑙=0
First, we can get 𝑢1 in order 𝑂(𝜏2 + ℎ2 ) that satisfies 𝐵0 =
𝑂(𝜏2 + ℎ2 )2 by two-level 𝐶-𝑁 scheme. Since
󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2
≤ 𝐶 [󵄩󵄩󵄩𝑒𝑛𝑥 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑒𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑒𝑛 󵄩󵄩󵄩 ]
𝑛
2
󵄩 󵄩2
󵄩 󵄩2
𝜏∑󵄩󵄩󵄩󵄩𝑟𝑙 󵄩󵄩󵄩󵄩 ≤ 𝑛𝜏 max󵄩󵄩󵄩󵄩𝑟𝑙 󵄩󵄩󵄩󵄩 ≤ 𝑇 ⋅ 𝑂(𝜏2 + ℎ2 ) ,
󵄩2 󵄩
󵄩2 󵄩 󵄩2 󵄩
󵄩2 󵄩
󵄩2
󵄩
≤ 𝐶 [󵄩󵄩󵄩󵄩𝑒𝑛+1 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩𝑒𝑛−1 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑒𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩𝑒𝑥𝑛+1 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩𝑒𝑥𝑛−1 󵄩󵄩󵄩󵄩 ] .
1≤𝑙≤𝑛
𝑙=1
(44)
(53)
then we obtain
𝑛−1
2
𝐵𝑛 ≤ 𝑂(𝜏2 + ℎ2 ) + 𝐶𝜏 ∑ 𝐵𝑙 .
Noting that
(54)
𝑙=0
⟨𝑟𝑛 , 2𝑒𝑛 ⟩ = ⟨𝑟𝑛 , 𝑒𝑛+1 + 𝑒𝑛−1 ⟩
󵄩2 󵄩
󵄩2
󵄩 󵄩2 1 󵄩
≤ 󵄩󵄩󵄩𝑟𝑛 󵄩󵄩󵄩 + [󵄩󵄩󵄩󵄩𝑒𝑛+1 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩𝑒𝑛−1 󵄩󵄩󵄩󵄩 ] ,
2
(45)
From Lemma 6 we get
2
𝐵𝑛 ≤ 𝑂(𝜏2 + ℎ2 ) ,
and from (42)–(45), we have
(55)
which implies that
󵄩2 󵄩
󵄩2
󵄩 𝑛+1 󵄩󵄩2 󵄩󵄩 𝑛−1 󵄩󵄩2
󵄩
󵄩󵄩 − 󵄩󵄩𝑒𝑥𝑥 󵄩󵄩 )
(󵄩󵄩󵄩󵄩𝑒𝑛+1 󵄩󵄩󵄩󵄩 − 󵄩󵄩󵄩󵄩𝑒𝑛−1 󵄩󵄩󵄩󵄩 ) + (󵄩󵄩󵄩󵄩𝑒𝑥𝑥
󵄩 󵄩
󵄩
󵄩
󵄩2 󵄩 󵄩2 󵄩
󵄩2
≤ 𝐶𝜏 [󵄩󵄩󵄩󵄩𝑒𝑛+1 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑒𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩𝑒𝑛−1 󵄩󵄩󵄩󵄩
󵄩󵄩 𝑛 󵄩󵄩
2
2
󵄩󵄩𝑒 󵄩󵄩 ≤ 𝑂 (𝜏 + ℎ ) ,
(46)
(56)
From (47) we have
󵄩󵄩 𝑛 󵄩󵄩
2
2
󵄩󵄩𝑒𝑥 󵄩󵄩 ≤ 𝑂 (𝜏 + ℎ ) .
󵄩2 󵄩 󵄩2 󵄩
󵄩2
󵄩
󵄩 󵄩2
+󵄩󵄩󵄩󵄩𝑒𝑥𝑛+1 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑒𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩𝑒𝑥𝑛−1 󵄩󵄩󵄩󵄩 ] + 2𝜏󵄩󵄩󵄩𝑟𝑛 󵄩󵄩󵄩 .
(57)
By Lemma 5 we obtain
Similar to the proof of (38), we have
󵄩󵄩 𝑛 󵄩󵄩
2
2
󵄩󵄩𝑒 󵄩󵄩∞ ≤ 𝑂 (𝜏 + ℎ ) .
󵄩󵄩 𝑛+1 󵄩󵄩2 1 󵄩󵄩 𝑛+1 󵄩󵄩2 󵄩󵄩 𝑛+1 󵄩󵄩2
󵄩󵄩𝑒𝑥 󵄩󵄩 ≤ (󵄩󵄩𝑒 󵄩󵄩 + 󵄩󵄩𝑒𝑥𝑥 󵄩󵄩 ) ,
󵄩
󵄩
󵄩 󵄩
󵄩
2 󵄩
󵄩󵄩 𝑛 󵄩󵄩2 1 󵄩󵄩 𝑛 󵄩󵄩2 󵄩󵄩 𝑛 󵄩󵄩2
󵄩󵄩𝑒𝑥 󵄩󵄩 ≤ (󵄩󵄩𝑒 󵄩󵄩 + 󵄩󵄩𝑒𝑥𝑥 󵄩󵄩 ) ,
2
󵄩󵄩 𝑛 󵄩󵄩
2
2
󵄩󵄩𝑒𝑥𝑥 󵄩󵄩 ≤ 𝑂 (𝜏 + ℎ ) .
(47)
(58)
Finally, we can similarly prove results as follows.
Theorem 9. Under the conditions of Theorem 8, the solution
𝑢𝑛 of (10)–(13) is stable in norm ‖ ⋅ ‖∞ .
󵄩󵄩 𝑛−1 󵄩󵄩2 1 󵄩󵄩 𝑛−1 󵄩󵄩2 󵄩󵄩 𝑛−1 󵄩󵄩2
󵄩󵄩𝑒𝑥 󵄩󵄩 ≤ (󵄩󵄩𝑒 󵄩󵄩 + 󵄩󵄩𝑒𝑥𝑥 󵄩󵄩 ) .
󵄩
󵄩 󵄩
󵄩
󵄩
2 󵄩
Then, (46) can be rewritten as
5. Numerical Simulations
󵄩2 󵄩
󵄩2
󵄩 𝑛+1 󵄩󵄩2 󵄩󵄩 𝑛−1 󵄩󵄩2
󵄩
󵄩󵄩 − 󵄩󵄩𝑒𝑥𝑥 󵄩󵄩 )
(󵄩󵄩󵄩󵄩𝑒𝑛+1 󵄩󵄩󵄩󵄩 − 󵄩󵄩󵄩󵄩𝑒𝑛−1 󵄩󵄩󵄩󵄩 ) + (󵄩󵄩󵄩󵄩𝑒𝑥𝑥
󵄩 󵄩
󵄩
󵄩
󵄩2 󵄩 󵄩2 󵄩
󵄩2
≤ 𝐶𝜏 [󵄩󵄩󵄩󵄩𝑒𝑛+1 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑒𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩𝑒𝑛−1 󵄩󵄩󵄩󵄩
(48)
󵄩 𝑛+1 󵄩󵄩2 󵄩󵄩 𝑛 󵄩󵄩2 󵄩󵄩 𝑛−1 󵄩󵄩2
󵄩󵄩 + 󵄩󵄩𝑒𝑥𝑥 󵄩󵄩 + 󵄩󵄩𝑒𝑥𝑥 󵄩󵄩 ] + 2𝜏󵄩󵄩󵄩󵄩𝑟𝑛 󵄩󵄩󵄩󵄩2 .
+󵄩󵄩󵄩󵄩𝑒𝑥𝑥
󵄩
󵄩
󵄩
2
2
2
2
𝑛+1
𝑛
Let 𝐵𝑛 = ‖𝑒𝑛+1 ‖ + ‖𝑒𝑥𝑥
‖ + ‖𝑒𝑛 ‖ + ‖𝑒𝑥𝑥
‖ . Then, (48) can be
rewritten as follows:
󵄩 󵄩2
𝐵𝑛 − 𝐵𝑛−1 ≤ 𝐶𝜏 (𝐵𝑛 + 𝐵𝑛−1 ) + 2𝜏󵄩󵄩󵄩𝑟𝑛 󵄩󵄩󵄩 ,
(49)
Since the three-level implicit finite difference scheme cannot
start by itself, we need to select other two-level schemes (such
as the 𝐶-𝑁 Scheme) to get 𝑢1 . Then, be reusing initial value
𝑢0 , we can work out 𝑢2 , 𝑢3 , . . .. Iterative numerical calculation
is not required, for this scheme is linear, so it saves computing
time. Let 𝑥𝐿 = −70, 𝑥𝑅 = 100, and 𝑇 = 40,
𝑢0 (𝑥) = (−
1
35 √
35
313) sech4 ( √−26 + 2√313𝑥) .
+
24 312
24
(59)
6
Advances in Mathematical Physics
Table 1: The error at various time steps.
𝜏 = ℎ = 0.1
‖𝑒𝑛 ‖∞
‖𝑒𝑛 ‖
1.641934𝑒 − 3
6.314193𝑒 − 4
3.045414𝑒 − 3
1.131442𝑒 − 3
4.241827𝑒 − 3
1.533771𝑒 − 3
5.297873𝑒 − 3
1.878952𝑒 − 3
𝑡 = 10
𝑡 = 20
𝑡 = 30
𝑡 = 40
𝜏 = ℎ = 0.05
‖𝑒𝑛 ‖
‖𝑒𝑛 ‖∞
4.113510𝑒 − 4
1.582641𝑒 − 4
7.631169𝑒 − 4
2.835874𝑒 − 4
1.062971𝑒 − 3
3.843906𝑒 − 4
1.327645𝑒 − 3
4.709118𝑒 − 4
𝜏 = ℎ = 0.025
‖𝑒𝑛 ‖
‖𝑒𝑛 ‖∞
1.028173𝑒 − 4
3.965867𝑒 − 5
1.905450𝑒 − 4
7.097948𝑒 − 5
2.650990𝑒 − 4
9.610332𝑒 − 5
3.306738𝑒 − 4
1.176011𝑒 − 4
Table 2: The verification of the second convergence.
‖𝑒𝑛 (ℎ, 𝜏)‖/‖𝑒2𝑛 (ℎ/2, 𝜏/2)‖
𝜏 = ℎ = 0.05
𝜏 = ℎ = 0.025
3.991564
4.000797
3.990757
4.004916
3.990539
4.009713
3.990427
4.014970
𝜏 = ℎ = 0.1
—
—
—
—
𝑡 = 10
𝑡 = 20
𝑡 = 30
𝑡 = 40
‖𝑒𝑛 (ℎ, 𝜏)‖∞ /‖𝑒2𝑛 (ℎ/2, 𝜏/2)‖∞
𝜏 = ℎ = 0.05
𝜏 = ℎ = 0.025
3.989657
3.990655
3.989749
3.995343
3.990136
3.999764
3.990030
4.004314
𝜏 = ℎ = 0.1
—
—
—
—
Table 3: Numerical simulations on the two conservation invariants 𝑄𝑛 and 𝐸𝑛 .
𝜏 = ℎ = 0.1
𝑛
𝑡=0
𝑡 = 10
𝑡 = 20
𝑡 = 30
𝑡 = 40
𝑛
𝐸
1.984553365290
1.984595075859
1.984645964099
1.984679827211
1.984701501262
𝑄
5.497722548019
5.497724936513
5.497728744900
5.497731963790
5.497734235191
𝑛
𝜏 = ℎ = 0.05
𝑄
5.498060684522
5.498060837192
5.498061080542
5.498061287046
5.498061398506
𝐸
1.984390175264
1.984401029470
1.984414367496
1.984423270337
1.984428974030
0.6
0.6
0.5
0.5
0.4
0.4
0.3
0.3
0.2
0.2
0.1
0.1
0
0
−0.1
−60
−40
−20
0
20
40
60
80
100
𝑡=0
𝑡 = 20
𝑡 = 40
𝑛
−0.1
−60
−40
−20
𝑛
𝜏 = ℎ = 0.025
𝑄
5.498145418391
5.498145479109
5.498145545374
5.498145609535
5.498145659050
0
20
40
𝐸𝑛
1.984349335263
1.984352109750
1.984355520610
1.984357811266
1.984359292230
60
80
100
𝑡=0
𝑡 = 20
𝑡 = 40
Figure 1: When 𝜏 = ℎ = 0.1, the wave graph of 𝑢(𝑥, 𝑡) at various
times.
Figure 2: When 𝜏 = ℎ = 0.025, the wave graph of 𝑢(𝑥, 𝑡) at various
times.
In Table 1, we give the error at various time steps. Using
the method in [30, 31], we verified the second convergence of
the difference scheme in Table 2. Numerical simulations on
two conservation invariants 𝑄𝑛 and 𝐸𝑛 are given in Table 3.
The wave graph comparison of 𝑢(𝑥, 𝑡) between 𝜏 = ℎ =
0.1 and 𝜏 = ℎ = 0.025 at various times is given in Figures 1
and 2.
Numerical simulations show that the finite difference
scheme is efficient.
Advances in Mathematical Physics
7
Acknowledgments
The work was supported by Scientific Research Fund of
Sichuan Provincial Education Department (11ZB009) and the
fund of Key Disciplinary of Computer Software and Theory,
Sichuan, Grant no. SZD0802-09-1.
[18]
[19]
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